Here's the problem I'm trying to solve:
If an odd function g(x) is right-continuous at x = 0, show that it is continuous at x = 0 and that g(0) = 0. Hint: Prove first that \(\displaystyle \lim_{x \to 0^{-}} g(x)\) exists and equals to \(\displaystyle \lim_{x \to 0^{+}} g(-x)\)
Here's my attempt:
I was thinking that since -g(x) = g(-x) then |-g(x)| = |g(-x)|. And since we're proving that the limit of odd continuous functions have a limit at 0 (I think that's a safe assumption) then:
Given \(\displaystyle \epsilon>0\). \(\displaystyle \exists\delta >0\) such that whenever 0 < |x| < \(\displaystyle \delta\) it follows that |g(x) - M| < \(\displaystyle \epsilon\). Since M = 0, then \(\displaystyle |g(x)|<\epsilon\). (All this is the function approaching 0 from the left)
Then looking at g(-x), for the same epsilon and delta (since the definition of odd functions implies symmetry about the origin) we arrive at the same conclusion |g(x)| < \(\displaystyle \epsilon\).
Do I seem to be on the right track or is this just dead wrong? It's difficult making the transition from concrete, repetitive epsilon-delta proofs to more general and abstract ones. The methods don't quite seem to be the same ...
Thanks a lot in advance! Any help would be deeply appreciated : )
If an odd function g(x) is right-continuous at x = 0, show that it is continuous at x = 0 and that g(0) = 0. Hint: Prove first that \(\displaystyle \lim_{x \to 0^{-}} g(x)\) exists and equals to \(\displaystyle \lim_{x \to 0^{+}} g(-x)\)
Here's my attempt:
I was thinking that since -g(x) = g(-x) then |-g(x)| = |g(-x)|. And since we're proving that the limit of odd continuous functions have a limit at 0 (I think that's a safe assumption) then:
Given \(\displaystyle \epsilon>0\). \(\displaystyle \exists\delta >0\) such that whenever 0 < |x| < \(\displaystyle \delta\) it follows that |g(x) - M| < \(\displaystyle \epsilon\). Since M = 0, then \(\displaystyle |g(x)|<\epsilon\). (All this is the function approaching 0 from the left)
Then looking at g(-x), for the same epsilon and delta (since the definition of odd functions implies symmetry about the origin) we arrive at the same conclusion |g(x)| < \(\displaystyle \epsilon\).
Do I seem to be on the right track or is this just dead wrong? It's difficult making the transition from concrete, repetitive epsilon-delta proofs to more general and abstract ones. The methods don't quite seem to be the same ...
Thanks a lot in advance! Any help would be deeply appreciated : )
